7.6 Solving systems with gaussian elimination (Page 4/13)

College algebra

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Solving a dependent system of linear equations using matrices

Solve the following system of linear equations using matrices.

\begin{array}{r} - x −2 y + z = −1 \\ 2 x + 3 y = 2 \\ y −2 z = 0 \end{array}

Write the augmented matrix.

[\begin{array}{r} −1 & −2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & −2 \end{array} | \begin{array}{r} −1 \\ 2 \\ 0 \end{array}]

First, multiply row 1 by $−1$ to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form.

- R_{1} \to [\begin{array}{r} 1 & 2 & −1 & 1 \\ 2 & 3 & 0 & 2 \\ 0 & 1 & −2 & 0 \end{array}]

R_{2} \leftrightarrow R_{3} \to [\begin{array}{r} 1 & 2 & - 1 \\ 0 & 1 & - 2 \\ 2 & 3 & 0 \end{array} | \begin{array}{r} 1 \\ 0 \\ 2 \end{array}]

−2 R_{1} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 2 & −1 \\ 0 & 1 & −2 \\ 0 & −1 & 2 \end{array} | \begin{array}{r} 1 \\ 0 \\ 0 \end{array}]

R_{2} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 2 & −1 \\ 0 & 1 & −2 \\ 0 & 0 & 0 \end{array} | \begin{array}{r} 2 \\ 1 \\ 0 \end{array}]

The last matrix represents the following system.

\begin{array}{l} x + 2 y - z = 1 \\ y - 2 z = 0 \\ 0 = 0 \end{array}

We see by the identity $0 = 0$ that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for $y$ and substituting it into the first equation we can solve for $z$ in terms of $x .$

\begin{array}{l} x + 2 y - z = 1 \\ y = 2 z \\ x + 2 (2 z) - z = 1 \\ x + 3 z = 1 \\ z = \frac{1 - x}{3} \end{array}

Now we substitute the expression for $z$ into the second equation to solve for $y$ in terms of $x .$

\begin{array}{l} y - 2 z = 0 \\ z = \frac{1 - x}{3} \\ y - 2 (\frac{1 - x}{3}) = 0 \\ y = \frac{2 - 2 x}{3} \end{array}

The generic solution is $(x, \frac{2 −2 x}{3}, \frac{1 - x}{3}) .$

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Try It

Solve the system using matrices.

\begin{matrix} x + 4 y - z = 4 \\ 2 x + 5 y + 8 z = 15 \\ x + 3 y −3 z = 1 \end{matrix}

$(1, 1, 1)$

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Q&A

Can any system of linear equations be solved by Gaussian elimination?

Yes, a system of linear equations of any size can be solved by Gaussian elimination.

How To

Given a system of equations, solve with matrices using a calculator.

Save the augmented matrix as a matrix variable $[A], [B], [C], \dots .$
Use the ref( function in the calculator, calling up each matrix variable as needed.

Solving systems of equations with matrices using a calculator

Solve the system of equations.

\begin{array}{r} 5 x + 3 y + 9 z = −1 \\ −2 x + 3 y - z = −2 \\ - x −4 y + 5 z = 1 \end{array}

Write the augmented matrix for the system of equations.

[\begin{array}{r} 5 & 3 & 9 \\ −2 & 3 & −1 \\ −1 & −4 & 5 \end{array} | \begin{array}{r} 5 \\ −2 \\ −1 \end{array}]

On the matrix page of the calculator, enter the augmented matrix above as the matrix variable $[A] .$

[A] = [\begin{array}{r} 5 & 3 & 9 & −1 \\ −2 & 3 & −1 & −2 \\ −1 & −4 & 5 & 1 \end{array}]

Use the ref( function in the calculator, calling up the matrix variable $[A] .$

ref ([A])

Evaluate.

\begin{array}{l} [\begin{array}{r} 1 & \frac{3}{5} & \frac{9}{5} & \frac{1}{5} \\ 0 & 1 & \frac{13}{21} & - \frac{4}{7} \\ 0 & 0 & 1 & - \frac{24}{187} \end{array}] \to \begin{array}{l} x + \frac{3}{5} y + \frac{9}{5} z = - \frac{1}{5} \\ y + \frac{13}{21} z = - \frac{4}{7} \\ z = - \frac{24}{187} \end{array} \end{array}

Using back-substitution, the solution is $(\frac{61}{187}, - \frac{92}{187}, - \frac{24}{187}) .$

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Applying 2 × 2 matrices to finance

Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?

We have a system of two equations in two variables. Let $x =$ the amount invested at 10.5% interest, and $y =$ the amount invested at 12% interest.

\begin{array}{l} x + y = 12,000 \\ 0.105 x + 0.12 y = 1,335 \end{array}

As a matrix, we have

[\begin{array}{r} 1 & 1 \\ 0.105 & 0.12 \end{array} | \begin{array}{r} 12,000 \\ 1,335 \end{array}]

Multiply row 1 by $−0.105$ and add the result to row 2.

[\begin{array}{r} 1 & 1 \\ 0 & 0.015 \end{array} | \begin{array}{r} 12,000 \\ 75 \end{array}]

Then,

\begin{array}{l} 0.015 y = 75 \\ y = 5,000 \end{array}

So $12,000 −5,000 = 7,000.$

Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.

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Applying 3 × 3 matrices to finance

Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?

We have a system of three equations in three variables. Let $x$ be the amount invested at 5% interest, let $y$ be the amount invested at 8% interest, and let $z$ be the amount invested at 9% interest. Thus,

\begin{array}{l} x + y + z = 10, 000 \\ 0.05 x + 0.08 y + 0.09 z = 770 \\ 2 x - z = 0 \end{array}

As a matrix, we have

[\begin{array}{r} 1 & 1 & 1 \\ 0.05 & 0.08 & 0.09 \\ 2 & 0 & −1 \end{array} | \begin{array}{r} 10, 000 \\ 770 \\ 0 \end{array}]

Now, we perform Gaussian elimination to achieve row-echelon form.

\begin{array}{l} \begin{array}{l} −0.05 R_{1} + R_{2} = R_{2} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 2 & 0 & −1 \end{array} | \begin{array}{r} 10,000 \\ 270 \\ 0 \end{array}] \end{array} \\ −2 R_{1} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 0.03 & 0.04 \\ 0 & −2 & −3 \end{array} | \begin{array}{r} 10,000 \\ 270 \\ −20,000 \end{array}] \\ \frac{1}{0.03} R_{2} = R_{2} \to [\begin{array}{r} 0 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & −2 & −3 \end{array} | \begin{array}{r} 10,000 \\ 9,000 \\ −20,000 \end{array}] \\ 2 R_{2} + R_{3} = R_{3} \to [\begin{array}{r} 1 & 1 & 1 \\ 0 & 1 & \frac{4}{3} \\ 0 & 0 & - \frac{1}{3} \end{array} | \begin{array}{r} 10,000 \\ 9,000 \\ −2,000 \end{array}] \end{array}

The third row tells us $- \frac{1}{3} z = −2,000;$ thus $z = 6,000.$

The second row tells us $y + \frac{4}{3} z = 9,000.$ Substituting $z = 6,000,$ we get

\begin{array}{r} y + \frac{4}{3} (6,000) = 9,000 \\ y + 8,000 = 9,000 \\ y = 1,000 \end{array}

The first row tells us $x + y + z = 10, 000.$ Substituting $y = 1, 000$ and $z = 6, 000,$ we get

\begin{array}{l} x + 1, 000 + 6, 000 = 10,000 \\ x = 3,000 \end{array}

The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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